A family with 5 members (Father, Mother, two daughters...

[BTGmoderatorLU]

A family with 5 members (Father, Mother, two daughters and one son) plan to go on a trip in car which has 5 seats (2 sets in front including driving seat and 3 seats at the back).

If all family members randomly sit on the the seats in car then what's the probability of an acceptable seating arrangement if two daughters do not want to sit next to each other and only one of the parents can drive?

A)1/120
B)1/60
C) 1/20
D) 4/15
E) 1/5

The OA is D.

Can any expert explain this PS question please? I don't have it clear.

[[email protected]]

Hi LUANDATO,

These types of questions can be approached a couple of different ways. There's a "visual" aspect to this question that can help you to take advantage of some shortcuts built into the prompt, so I'm going to use a bit of "brute force" and some pictures to answer this question. Since we're arranging people in seats, we'll end up doing some "permutation math."

To start, without any 'restrictions' there would be 5! = (5)(4)(3)(2)(1) = 120 possible arrangements. However, there ARE some restrictions that we have to follow....

M = Mother
F = Father
D1 = 1st Daughter
D2 = 2nd Daughter
S = Son

Front Back
_ _ _ _ _
1st spot = driver

We have 2 restrictions that we have to follow:
1) Either the Father or Mother must be the driver
2) The two daughters CANNOT sit next to one another

Let's put the Mother in the driver's seat and count up the possibilities:

M F (2)(1)(1) Here, the two daughters have to be separated by the son, but either daughter could be in the "first back seat" = 2 options

M D1 (3)(2)(1) Here, with the first daughter up front, the remaining 3 people (F, D2 and S) can be in any of the back seats = 6 options
M D2 (3)(2)(1) Here, we have the same situation, but with the second daughter up front... = 6 options
M S (2)(1)(1) Here, with the son up front, we have the same scenario as we had when the Father was up front = 2 options

Total options with Mother driving = 2+6+6+2 = 16 options

Now we can take advantage of the shortcut I mentioned earlier - We can flip-flop the Mother and Father in the above examples. This will gives us another 16 options with the Father driving.

Total options: 16 + 16 = 32 options

Thus, the probability of the 5 people sitting in such a way that 'fits' all the given restrictions is 32/120 = 4/15

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[[email protected]]

Hi LUANDATO,

These types of questions can be approached a couple of different ways. There's a "visual" aspect to this question that can help you to take advantage of some shortcuts built into the prompt, so I'm going to use a bit of "brute force" and some pictures to answer this question. Since we're arranging people in seats, we'll end up doing some "permutation math."

To start, without any 'restrictions' there would be 5! = (5)(4)(3)(2)(1) = 120 possible arrangements. However, there ARE some restrictions that we have to follow....

M = Mother
F = Father
D1 = 1st Daughter
D2 = 2nd Daughter
S = Son

Front Back
_ _ _ _ _
1st spot = driver

We have 2 restrictions that we have to follow:
1) Either the Father or Mother must be the driver
2) The two daughters CANNOT sit next to one another

Let's put the Mother in the driver's seat and count up the possibilities:

M F (2)(1)(1) Here, the two daughters have to be separated by the son, but either daughter could be in the "first back seat" = 2 options

M D1 (3)(2)(1) Here, with the first daughter up front, the remaining 3 people (F, D2 and S) can be in any of the back seats = 6 options
M D2 (3)(2)(1) Here, we have the same situation, but with the second daughter up front... = 6 options
M S (2)(1)(1) Here, with the son up front, we have the same scenario as we had when the Father was up front = 2 options

Total options with Mother driving = 2+6+6+2 = 16 options

Now we can take advantage of the shortcut I mentioned earlier - We can flip-flop the Mother and Father in the above examples. This will gives us another 16 options with the Father driving.

Total options: 16 + 16 = 32 options

Thus, the probability of the 5 people sitting in such a way that 'fits' all the given restrictions is 32/120 = 4/15

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

A family with 5 members (Father, Mother, two daughters and one son) plan to go on a trip in car which has 5 seats (2 sets in front including driving seat and 3 seats at the back).

If all family members randomly sit on the the seats in car then what's the probability of an acceptable seating arrangement if two daughters do not want to sit next to each other and only one of the parents can drive?

A)1/120
B)1/60
C) 1/20
D) 4/15
E) 1/5

Constraints:
A parent must sit in the driver's seat.
The 2 daughters must sit in nonadjacent seats.

In probability problems:
AND means MULTIPLY.
OR means ADD.

Case 1: A parent in the driver's seat and a daughter in the front passenger seat
P(a parent in the driver's seat) = 2/5. (Of the 5 passengers, 2 are parents.)
P(a daughter in the front passenger seat) = 2/4. (Of the 4 remaining passengers, 2 are daughters.)
Since we want a parent in the driver's seats AND a daughter in the front passenger seat, we MULTIPLY the fractions:
2/5 * 2/4 = 1/5.

Case 2: A parent in the driver's seat and the 2 daughters in the left and right back passenger seats
P(a parent in the driver's seat) = 2/5. (Of the 5 passengers, 2 are parents.)
P(a daughter in the left back seat) = 2/4. (Of the 4 remaining passengers, 2 are daughters.)
P(a daughter in the right back seat) = 1/3. (Of the 3 remaining passengers, 1 is a daughter.)
Since we want a parent in the driver's seat AND a daughter in the left back seat AND the other daughter in the right back seat, we MULTIPLY the fractions:
2/5 * 2/4 * 1/3 = 1/15.

Since a good outcome will be yielded by Case 1 OR Case 2, we ADD the fractions in blue:
1/5 + 1/15 = 3/15 + 1/15 = 4/15.

The correct answer is D.

[Scott@TargetTestPrep]

A family with 5 members (Father, Mother, two daughters and one son) plan to go on a trip in car which has 5 seats (2 sets in front including driving seat and 3 seats at the back).

If all family members randomly sit on the the seats in car then what's the probability of an acceptable seating arrangement if two daughters do not want to sit next to each other and only one of the parents can drive?

A)1/120
B)1/60
C) 1/20
D) 4/15
E) 1/5

The OA is D.

Can any expert explain this PS question please? I don't have it clear.

There are a total of 5! = 120 seating arrangements, acceptable or not.

Let's suppose father is the driver. Without the condition on the daughters, there are 4! = 24 ways the remaining four people can be seated. However, we have the following unacceptable seating arrangements for the back seat:

D1-D2-X, D2-D1-X, X-D1-D2, X-D2-D1.

In the above seating arrangements, there are two choices for X; thus there are a total of 4 x 2 = 8 unacceptable seating arrangements. Therefore, when father is the driver, there are 24 - 8 = 16 acceptable seating arrangements.

We note that all of the above will also be valid when mother is the driver; thus, there are another 16 acceptable seating arrangements when mother is the driver.

Since there are a total of 16 + 16 = 32 acceptable seating arrangements, the probability that a random seating arrangement will be acceptable is 32/120 = 4/15.

Answer: D